I want to prove that there is a bijection between any interval in the rationals and all the rationals. I'll try to prove a special case and I'm hoping to know if this proof is correct. This is not a homework problem, it is merely a curiosity. I'm not so experienced in math proofs.
I need to prove a bijection exists between any enumeration of all rationals in $(0,1]$ and any enumeration of the rationals in $(0,M]$.
If $s_n$ is an enumeration of the set of all rational numbers in the interval $(0,M]$ where $M$ is a rational number, then:
I am able to define $u_n = s_n/M$ with the property that $0 < u_n \le $. $u_n$ is an enumeration of all the rationals between $0$ and $M$ because if there were any empty space $w$, where $0 < w \le 1$ and $\forall n, u_n \ne w$ it would imply that $Mw$ is not in the set $s_n$ which is a contradiction.
Moreover the interval $(0,1]$ is wholly mappable onto any interval $(0,M]$ since any enumeration of $(0,1]$ multiplied by $M$ is a subset of $(0,M]$. Assume there is an empty space $u$ in the interval $(0,M]$. But $u/M \in (0,1]$. So there are no empty spaces.
Since a map exists in both directions a bijection exists between all the rationals and the subset $(0,1]$ from which I am able to prove that if a bijection exists between $A$ and $(0,1]$ and so too for $B$ then a bijection exists between $A$ and $B$.
If this is true I would be very excited.
The problem with your proof is that if $M$ is itself not a rational number, then $s_n/M$ will not be a rational number.
What you can do instead is note that the rationals in $(0,1]$ and the rationals in $(0,M]$ are both countably infinite, so there are enumerations $(q_n)_n$ and $(r_n)_n$ enumerating them respectively. Then the map sending $q_n$ to $r_n$ for each $n$ is a bijection between the two sets of rationals in question.